Question

我想查找列值以数字开头的所有行。

它适用于此请求：

    WHERE trim(u_ods_val3.ods_itn_PHRSBMO.NO_ART_TECH_OI)    IS NOT NULL
  AND (SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)='0'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='1'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='2 '
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='3'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='4'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='5'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='6'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='7'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='8'
  OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)  ='9')

但它太长了。

感谢您的帮助。

Answer 1

Regexp_like会派上用场，而且会更短

where regexp_like(trim(col_name), '^[0-9]')

或使用字符类

where regexp_like(trim(col_name), '^[[:digit:]]')

Answer 2

尝试使用in：

WHERE trim(u_ods_val3.ods_itn_PHRSBMO.NO_ART_TECH_OI)    IS NOT NULL
  AND SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) in ('0','1','2','3','4','5','6','7','8','9')

Answer 3

BETWEEN就是您所需要的！（NOT NULL在这种情况下是隐含的！）

WHERE SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) between '0' and '9'

如果您在该列上有索引，并且不介意此解决方案的小脏点，您甚至可以加快它的速度：

WHERE u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI between '0' and '9~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'

这假定NO_ART_TECH_OI不包含带有ascii代码的字符＆gt; 126。

Oracle SQL：以数字开头的搜索列

3 个答案: